Posted by Serenity on .
Please help with this question, I have no clue how to go about it.
1.3188g of antacid is weighed and mixed with 75.00mL of excess 0.1746 M HCl. The excess acid required 27.20 mL of 0.09767 M NaOH for back titration. Calculate the neutralizing power of the antacid in terms of mmol H+ per gram of antacid.
mmoles HCl added = 0.1746 x 75.00 mL = ??
mmoles NaOH to titrate the excess acid = 27.20 mL x 0.09767 M NaOH = ?? moles NaOH.
The difference is the mmoles of the antacid in the 1.3188 g. That difference divided by the 1.3188 will be what you are looking for.
mmoles HCl added= 0.1746 x 75.00mL= 13.095 moles
mmles NaOH titrate the excess acid= 27.20 mL x 0.09767 M NaOH= 2.6566 moles
But I don't understand how to find the mmoles in 1.3188g.
Serenity-To Dr Bob222,
I also don't understand what you meant by the difference is the mmoles of the antacid in the 1.3188 g.
Please help, when possible.